Question

# Ten persons, amongst whom are $$A,B$$ and $$C$$ are to speak at a function. The number of ways in which it can be done if $$A$$ wants to speak before $$B$$, and $$B$$ wants to speak before $$C$$ is

A
10!6
B
8!
C
8!3
D
3!

Solution

## The correct option is D $$\displaystyle\frac{10!}{6}$$Places for A, B and C can be chosen in $$^{10}C_3$$ waysRemaining 7 can be selected in $$7!$$ waysSo total $$= ^{10}C_3\times 7!$$$$=\displaystyle \frac{10!.7!}{3!.7!}$$$$=\displaystyle \frac{10!}{6}$$Hence, option 'A' is correct.Maths

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